Constructing generator matrix of a linear code 
The linear code $C \cong \mathbb{F}^5_2$ is given by $C = \{(x_1, x_2, x_3, x_4, x_5) | x_1 + x_2 + x_3 = 0, x_4 + x_5 = 0$ in $\mathbb{F}_2\}$.
Write down a parity check matrix and a generator matrix for $C$.

For the parity check matrix I've let $\underline{x} = (x_1, x_2, x_3, x_4, x_5)$.
So for the condition $ x_1 + x_2 + x_3 = 0$ we have $\underline{x}.(1 1 1 0 0)= 0$.
And for the condition  $x_4 + x_5 = 0$ we have $\underline{x}.(0 0 0 1 1)= 0$
So a parity check matrix is:
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
\end{bmatrix}
How do I then go on to construct a generator matrix? I'm struggling to understand my notes and don't know how to begin. I just know that the dimension of $C$ is $5 - 2 = 3$ so the generator matrix will have $5$ columns and $3$ rows.
 A: Once you have $H$, to find $G$ you need to find a set of $k$ rows of length $n$ that are LI and orthogonal to the rows of $H$. 
In a simple case like this, it could be done by trying, eg:
$$ G=
\begin{bmatrix}
1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$$
A more general and systematic way is to try to find an equivalent code (same codebook, different mapping) by doing elementary operations with the rows of $H$, to put it in systematic form: $H=( I | P´)$ and then $G=( P | I)$ fits the bill.
In this case, we cannot do that manipulating the rows alone. We can resort to an aditional trick: you are also allowed to permute some columns , to bring it to systematic form, but then at the end un-permute the rows in the resulting $G$.
So, let's permute columns 2 and 4 to get the modified parity matrix
$$\begin{bmatrix}
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}=[I | P']
$$
and the modified generator matrix is
$$[P | I] =\begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 \\
\end{bmatrix}
$$
and after permuting columns 2 and 4 we get
$$ G = \begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1
\end{bmatrix}
 $$
You can (you should) check that the rows are indeed LI and orthogonal to the rows of $H$.
